Chapter 10, Problem 54P

(a)

To determine

The rotational kinetic energy.

Answer to Problem 54P

The rotational kinetic energy for the system is $6.90\text{ J}$.

Explanation of Solution

Redraw the figure P10.54.

Consider that the vertically standing to be initial position and horizontal to be the final position.

Write the equation for conservation of energy.

  $\Delta K=\Delta U$

Here, $\Delta K$ is the change in kinetic energy and $\Delta U$ is the change in potential energy.

From the law of conservation of energy, gain in rotational kinetic energy equals to loss in gravitational potential energy for the given system.

Write the expression for rotational kinetic energy.

    $K_{\text{rotational}}=U_{\text{sphere}}+U_{\text{rod}}$                                                                                  (I)

Here, $K_{\text{rotational}}$ is rotational kinetic energy, $U_{\text{sphere}}$ is loss in gravitational potential energy for sphere and $U_{\text{rod}}$ is loss in gravitational potential energy for rod.

Write the expression for loss in gravitational potential energy for sphere.

    $U_{\text{sphere}}=Mg\left(l+\frac{d}{2}\right)$

Here, $M$ is the mass of sphere, $l$ is the length of rod, $g$ is the acceleration under gravity and $d$ is the diameter of sphere.

Write the expression for loss in gravitational potential energy for rod.

    $U_{\text{rod}}=mg\left(\frac{l}{2}\right)$

Here, $m$ is the mass of rod.

Substitute $Mg\left(l+\frac{d}{2}\right)$ for $U_{\text{sphere}}$ and $mg\left(\frac{l}{2}\right)$ for $U_{\text{rod}}$ in equation (I)

    $K_{\text{rotational}}=Mg\left(l+\frac{d}{2}\right)+mg\left(\frac{l}{2}\right)$                                                                  (II)

Conclusion:

Substitute $24.0\text{ cm}$ for $l$, $1.20\text{ kg}$ for $m$, $8.00\text{ cm}$ for $d$, $2.00\text{ kg}$ for $M$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (II).

    $\begin{array}{c}{K}_{\text{rotational}}=\left[\begin{array}{l}2.00\text{ kg}\left(9.80{\text{ m/s}}^2\right)\left(24.0\text{ cm}+\frac{8.00\text{cm}}{2}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ +1.20\text{ kg}\left(9.80{\text{ m/s}}^2\right)\left(\frac{24.0\text{ cm}}{2}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\end{array}\right]\\ =\left[\left(19.6\text{ kg}\cdot {\text{m/s}}^2\right)\left(0.28\text{m}\right)+\left(11.76\text{ kg}\cdot {\text{m/s}}^2\right)\left(0.12\text{m}\right)\right]\\ =5.488\text{ J+1}\text{.4112 J}\\ \text{=6}\text{.90 J}\end{array}$

Thus, the rotational kinetic energy for the system is $6.90\text{ J}$.

(b)

To determine

The angular speed of the rod and ball.

Answer to Problem 54P

The angular speed of the ball and the rod is $8.73\text{ rad/s}$.

Explanation of Solution

Write the expression for moment of inertia of sphere at center.

    $I_{SA}=\frac{2}{5}M{R}^2$

Here, $I_{SA}$ is moment of inertia of sphere at point A.

Write the expression for the parallel axis theorem for moment of inertia at point $O$.

    $I_{SO}=I_{SA}+M{\left(D\right)}^2$

Here, $I_{SO}$ is moment of inertia at point $O$ and $D$ is the distance from centre.

Substitute $\frac{2}{5}M{R}^2$ for $I_{SA}$, $l+\frac{d}{2}$ for $D$ and $\frac{d}{2}$ for $R$ in above equation and simplify.

    $I_{SO}=\frac{M{d}^2}{10}+M{\left(l+\frac{d}{2}\right)}^2$

Write the expression for moment of inertia of rod at point $O$.

    $I_{RO}=\frac{1}{3}m{l}^2$

Here, $I_{RO}$ is moment of inertia of rod at point $O$.

Write the expression for net moment of inertia for the whole system.

    $I_O=I_{SO}+I_{RO}$

Here, $I_O$ is the net moment of inertia for the whole system.

Substitute $\frac{M{d}^2}{10}+M{\left(l+\frac{d}{2}\right)}^2$ for $I_{SO}$ and $\frac{1}{3}m{l}^2$ for $I_{RO}$ in above equation.

    $I_O=\frac{M{d}^2}{10}+M{\left(l+\frac{d}{2}\right)}^2+\frac{1}{3}m{l}^2$                                                                 (III)

Write the expression for rotational kinetic energy.

    $K_{rotational}=\frac{1}{2}{I}_O{\omega }^2$

Here, $\omega$ is the angular velocity and $K_{rotational}$ is the rotational kinetic energy.

Simply the above equation for value of $\omega$.

    $\omega =\sqrt{\frac{2K_{\text{rotational}}}{I_O}}$                                                                                          (IV)

Conclusion:

Substitute $24.0\text{ cm}$ for $l$, $1.20\text{ kg}$ for $m$, $8.00\text{ cm}$ for $d$ and $2.00\text{ kg}$ for $M$ in equation (III).

$\begin{array}{c}{I}_O=\left[\begin{array}{l}\frac{2.00\text{ kg}{\left(8.00\text{ cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{10}\\ +2.00\text{ kg}{\left[\left(24.0\text{ cm}+\frac{8.00\text{ cm}}{2}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right]}^2\\ +\frac{1}{3}\left(1.20\text{ kg}\right){\left(24.0\text{ cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\end{array}\right]\\ =0.2\text{ kg}{\left(0.08\text{m}\right)}^2+2.00\text{ kg}{\left(\text{0}\text{.28}\text{m}\right)}^2+0.4\text{ kg}{\left(0.24\text{m}\right)}^2\\ =1.28\times {10}^{-3}\text{ kg}\cdot {\text{m}}^2+0.1568\text{ kg}\cdot {\text{m}}^2+0.02304\text{ kg}\cdot {\text{m}}^2\\ =0.18112\text{ kg}\cdot {\text{m}}^2\end{array}$

Substitute $0.18112\text{ kg}\cdot {\text{m}}^2$ for $I_O$ and $6.90\text{ J}$ for $K_{\text{rotational}}$ in equation (IV).

    $\begin{array}{c}\omega =\sqrt{\frac{2\left(6.90\text{ J}\right)}{0.18112{\text{ kg-m}}^2}}\\ =\sqrt{76.1925}\\ =8.73\text{ rad/s}\end{array}$

Thus, the angular speed of the ball and the rod is $8.73\text{ rad/s}$.

(c)

To determine

Thelinear speed of the center of mass of the ball.

Answer to Problem 54P

The linear speed of the ball of center of mass is $2.44\text{ m/s}$.

Explanation of Solution

Write the expression for linear speed of the ball.

    $v=r\omega$

Here, $v$ is the linear speed and $r$ is the length of rod and sphere.

Substitute $l+\frac{d}{2}$ for value of $r$ in the above equation.

    $v=\left(l+\frac{d}{2}\right)\omega$ (V)

Here, $v$ is the linear speed.

Conclusion:

Substitute $8.73\text{ rad/s}$ for $\omega$, $24.0\text{ cm}$ for $l$ and $8.00\text{ cm}$ for $d$ in equation (V)

    $\begin{array}{c}v=\left(24.0\text{ cm}+\frac{8.00\text{ cm}}{2}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(8.73\text{ rad/s}\right)\\ \text{=}\left(0.28\text{m}\right)\left(8.73\text{ rad/s}\right)\\ \text{=2}\text{.44 m/s}\end{array}$

Thus, the linear speed of the ball of center of mass is $2.44\text{ m/s}$.

(d)

To determine

Compare the speed with the speed had the ball fallen freelythrough the same distance of $28.0\text{ cm}$.

Answer to Problem 54P

The rod pulls the sphere down together while rotating by the speed factor $1.0432\text{ times}$.

Explanation of Solution

Loss in gravitational potential energy will be equal to gain in kinetic energy.

Write the expression for the conservation of energy.

    $\Delta K=\Delta U$                                                                                                   (VI)

Write the expression for loss in gravitational potential energy for sphere.

    $\Delta U_{\text{sphere}}=Mg\left(l+\frac{d}{2}\right)$

Here, $\Delta U_{sphere}$ is the change in potential energy of sphere.

Write the expression for gain kinetic energy.

    $\Delta K=\frac{1}{2}M{v}_{new}{}^2$

Here, $\Delta K$ is the change in kinetic energy and $v_{new}$ is new velocity for the system.

Substitute $\frac{1}{2}M{v}_{new}{}^2$ for $\Delta K$ and $Mg\left(l+\frac{d}{2}\right)$ for $\Delta U_{sphere}$ in equation (VI) and solve for $v_{new}$.

    $v_{\text{new}}=\sqrt{2g\left(l+\frac{d}{2}\right)}$                                                                                                    (VII)

Write the expression for the ratio of new speed to the original speed.

    $k=\frac{v_{\text{new}}}{v}$ (VIII)

Here, $k$ is the factor by which the rod pulls the sphere down.

Conclusion:

Substitute $9.80{\text{ m/s}}^2$ for $g$, $24.0\text{ cm}$ for $l$ and $8.00\text{ cm}$ for $d$ in equation (VIII)

    $\begin{array}{c}{v}_{\text{new}}=\sqrt{2\left(9.80{\text{ m/s}}^2\right)\left(24.0\text{ cm}+\frac{8.00\text{ cm}}{2}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)}\\ =\sqrt{\left(19.6{\text{ m/s}}^2\right)\left(\text{0}\text{.28}\text{m}\right)}\\ =\sqrt{5.488{\text{ m}}^2{\text{/s}}^{\text{2}}}\\ =2.34\text{ m/s}\end{array}$

Substitute $2.34\text{ m/s}$ for $v_{new}$ and $2.44\text{ m/s}$ for $v$ in equation (VIII).

    $\begin{array}{c}k=\frac{2.44}{2.34}\\ =1.0432\end{array}$

Thus, the rod pulls the sphere down together while rotating by more speed than in direct falling by the factor of  $1.0432\text{ times}$.